In the given reaction $Cd$ is being oxidized to $Cd^{+2}$ ions and $Ag^+$ ions are being reduced to Ag metal.Thus the galvanic cell in which the given reaction take place can be represented as $Cd(s)/Cd^{+2}(aq)\parallel Ag^+(aq)/Ag(s)$
Thus the standard potential of given reaction would be standard emf of above cell $(E^0_{cell})$ which is given cell
$E^0_{cell}=E^0_{cathode}-E^0_{Anode}$
$\Rightarrow E^0_{Ag^+/Ag}-E^0_{Cd^{+2}/cd}$
$\Rightarrow 0.8-0.4$
$\Rightarrow 0.4V$
Hence (b) is the correct answer.